{"id":247,"date":"2017-08-21T04:49:16","date_gmt":"2017-08-21T04:49:16","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=247"},"modified":"2020-02-03T18:01:27","modified_gmt":"2020-02-03T18:01:27","slug":"writing-names-and-formulas","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/writing-names-and-formulas\/","title":{"rendered":"Writing Names and Formulas"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/comparison-of-properties\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-naming-and-structure\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/acids-and-oxides\">Next Lesson \u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Chemical Naming and Structure: Writing Names and Formulas<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, we will review how oxidation numbers are assigned. We will also review how to write chemical names and formulas for molecular compounds, ionic compounds, and acids.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Ionic compounds are held together by strong ionic bonds, while molecular compounds are held together by much weaker intermolecular attractions\u2014either van der Waals attractions or hydrogen bonds.<\/li>\n<li>Substances with weaker attractions vaporize more easily, giving rise to higher vapor pressure.<\/li>\n<li>Substances with weaker attractions can be moved out of position more easily, leading to a lower melting point.<\/li>\n<li>Substances with weaker attractions can be separated from each other more easily, therefore having lower boiling points.<\/li>\n<\/ul>\n<section>\n<h3>Oxidation Numbers<\/h3>\n<p>There are situations in which an atom does not have a clear and unmistakable charge, yet it would be helpful to determine whether or not it has lost or gained electrons. This is particularly true when dealing with oxidation-reduction reactions. In these situations, we need <abbr title=\"The charge that an atom would have if all bonding electrons were assigned to the more electronegative element \">oxidation numbers<\/abbr>. By convention in this lesson, oxidation numbers are written with the sign given first, followed by the number. Charges are given in the reverse order. Thus, \u201c+2\u201d is an oxidation number, while \u201c2+\u201d is a charge.<\/p>\n<h3>Assigning Oxidation Numbers<\/h3>\n<p>Oxidation numbers may be assigned to individual atoms by following these simple rules:<\/p>\n<ul>\n<li>1. Elements have oxidation numbers of zero.<\/li>\n<li>2. Hydrogen in a compound is usually +1. The only notable exception would be a metal hydride, such as NaH, in which hydrogen would be -1.<\/li>\n<li>3. Oxygen in a compound is usually \u22122. Again, the most important exceptions would be peroxides, such as H<sub>2<\/sub>O<sub>2<\/sub> or BaO<sub>2<\/sub>, in which oxygen is \u22121.<\/li>\n<li>4. Monatomic ions have an oxidation number equal to their charge.<\/li>\n<li>5. All other oxidation numbers are assigned so that the sum of the oxidation numbers equals the charge on the species.<\/li>\n<\/ul>\n<h3>Examples<\/h3>\n<p>Rule 1 is straightforward. Fe, N<sub>2<\/sub>, and As are all examples of species whose oxidation numbers equal zero because they are elemental. Please remember that species such as Fe<sup>2+<\/sup> or N<sup>3\u2212<\/sup> are not elements but ions; thus, they do not fall under this rule.<\/p>\n<p>Rules 2 and 3 are also rather simple. In the water molecule (H<sub>2<\/sub>O), each hydrogen atom has an oxidation number of +1 and the oxygen atom is \u22122. In sulfurous acid (H<sub>2<\/sub>SO<sub>3<\/sub>), each hydrogen atom is still +1 and each oxygen atom is still \u22122.<\/p>\n<p>Rule 4 is easy when ions are written separately. The Cr<sup>3+<\/sup> ion has an oxidation number of +3 while the Cl<sup>1\u2212<\/sup> ion is \u22121. The use of this rule is less obvious when the ions are in a compound. For example, potassium sulfide (K<sub>2<\/sub>S) consists of a potassium ion, K+, and a sulfide ion, S<sup>2\u2212<\/sup>. We do not see these charges in the compound\u2019s formula, so we must infer them. Once we do, we can see that the oxidation numbers are +1 and \u22122, respectively.<\/p>\n<p>Rule 5 requires the most thought. We invoke this rule when none of the others apply. Let\u2019s reconsider H<sub>2<\/sub>SO<sub>3<\/sub>. The sulfur atom is not a simple sulfide ion here; rather, it is part of the sulfite ion, SO<sub>3<\/sub><sup>2\u2212<\/sup>. As such, we cannot assign it a \u22122 oxidation number. Instead, we remember that each hydrogen is +1 and each oxygen is \u22122. In addition, bear in mind that the compound as a whole has no charge. Thus, all oxidation numbers must add up to zero. Representing sulfur\u2019s unknown oxidation number with the variable \u201cx\u201d, we may find it with the following expression:<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s7_001.gif\" \/><\/center>Let\u2019s consider one more example\u2014the dichromate ion, Cr<sub>2<\/sub>O<sub>7<\/sub><sup>2\u2212<\/sup>. The oxidation number for each chromium is unknown, but each oxygen is still \u22122. Because we are dealing with an ion rather than a neutral compound, we now want the sum of oxidation numbers to equal \u22122. The following shows how chromium\u2019s oxidation number is found:<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s7_002.gif\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question 1<\/h4>\n<p>Which of the following gives the correct oxidation number for chlorine in calcium hypochlorite, Ca(ClO)<sub>2<\/sub><\/p>\n<ol>\n<li>\u22121<\/li>\n<li>+1<\/li>\n<li>+3<\/li>\n<li>+5<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B. Calcium is a monatomic ion (Ca<sup>2<\/sup><sup>+<\/sup>) so its oxidation number is +2. Each oxygen is \u22122. The compound has no net charge so all oxidation numbers must add up to zero.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 2<\/h4>\n<p>Which of the following compounds has the highest boiling point?<\/p>\n<ol>\n<li>Water, H<sub>2 <\/sub>O<\/li>\n<li>Sodium chloride, NaCl<\/li>\n<li>Sulfur dioxide, SO<sub>2<\/sub><\/li>\n<li>Methanol, CH<sub>3<\/sub>OH<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B. Calcium is a monatomic ion (Ca<sub>2<\/sub><sup>+<\/sup>) so its oxidation number is +2. Each oxygen is \u22122. The compound has no net charge so all oxidation numbers must add up to zero. Chlorine is assigned as follows:<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 3<\/h4>\n<p>Which of the following correctly assigns the oxidation number to iron in Fe<sub>2<\/sub>(CO<sub>3<\/sub>)<sub>3<\/sub>?<\/p>\n<ol>\n<li>+3<\/li>\n<li>+2<\/li>\n<li>+1<\/li>\n<li>0<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice A is correct. The iron exists as a monatomic ion in this compound; however, iron may only have a charge of 2+ or 3+. This eliminates choices C and D. Because the carbonate ion has a 2\u2212 charge and there are three of them, each of the two iron ions must have a 3+ charge for the compound to be electrically neutral. This means that the oxidation number is +3.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 4<\/h4>\n<p>In which of the following compounds does hydrogen not have an oxidation number of +1?<\/p>\n<ol>\n<li>Hydrogen peroxide, H<sub>2<\/sub>O<sub>2<\/sub><\/li>\n<li>Hydrogen fluoride, HF<\/li>\n<li>Cesium hydride, CsH<\/li>\n<li>Hydrogen is always +1 in a compound<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice C is correct. In a metal hydride, hydrogen is a monatomic ion with a charge of 1\u2212, therefore, its oxidation number is equal to the ion charge, \u22121.<\/p>\n<\/section>\n<h3>Molecular Names and Formulas<\/h3>\n<p>We recall that molecular compounds are formed when nonmetal atoms share valence electrons, frequently to achieve eight valence electrons for each atom. Often, the same two nonmetal atoms may be able to combine in more than one ratio, thus forming more than one potential compound. For example, sulfur is known to combine with oxygen in a 1:2 ratio (SO<sub>2<\/sub>) and in a 1:3 ratio (SO<sub>3<\/sub>). Since multiple combinations are possible, it is important that the name of a molecular compound indicates how many of each atom is present in that compound. This is done with prefixes. Below are listed the prefixes used in naming molecular compounds, along with the numerical value that each prefix represents.<\/p>\n<table>\n<thead>\n<tr>\n<th>Prefix<\/th>\n<th>Number<\/th>\n<th>Prefix<\/th>\n<th>Number<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>mono<\/td>\n<td>one<\/td>\n<td>hexa<\/td>\n<td>six<\/td>\n<\/tr>\n<tr>\n<td>di<\/td>\n<td>two<\/td>\n<td>hepta<\/td>\n<td>seven<\/td>\n<\/tr>\n<tr>\n<td>tri<\/td>\n<td>three<\/td>\n<td>octa<\/td>\n<td>eight<\/td>\n<\/tr>\n<tr>\n<td>tetra<\/td>\n<td>four<\/td>\n<td>nona<\/td>\n<td>nine<\/td>\n<\/tr>\n<tr>\n<td>penta<\/td>\n<td>five<\/td>\n<td>deca<\/td>\n<td>ten<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The name of a <abbr title=\" A compound made of only two nonmetal elements\">binary molecular compound<\/abbr> follows a straightforward pattern. The first element\u2019s name is listed first, preceded by whichever prefix indicates how many of that element\u2019s atoms are in the compound. The only exception to this is if there is only one of that atom. In that case, the prefix mono is omitted. The name of the second element follows, modified so that it ends in \u2013ide. It is always preceded by the appropriate prefix. A few examples are listed below:<\/p>\n<ul>\n<li>CS<sub>2\u00a0<\/sub>carbon disulfide<\/li>\n<li>PCl<sub>5\u00a0<\/sub>phosphorus pentachloride<\/li>\n<li>N<sub>2<\/sub>O<sub>4\u00a0<\/sub>dinitrogen tetroxide<\/li>\n<li>O<sub>2<\/sub>F<sub>2\u00a0<\/sub>dioxygen difluoride<\/li>\n<\/ul>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following correctly names the compound XeF4?<\/p>\n<ol>\n<li>Monoxenon tetrafluoride<\/li>\n<li>Monoxenon tetrafluorine<\/li>\n<li>Xenon tetrafluoride<\/li>\n<li>Xenon tetrafluorine<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C. Because there is only one xenon atom, the prefix mono is omitted, making choices A and B incorrect. The name of the second element in the compound must be modified so that it ends with \u2013ide, which was not done in choice D.<\/p>\n<\/section>\n<h3>Overview of Ionic Compounds<\/h3>\n<p>In ionic compounds, positive ions combine with negative ions in a ratio such that their charges add up to zero. This means that a particular pair of ions will always combine in the same ratio, which makes the use of prefixes unnecessary. For this reason, when we name any ionic compound, we simply name the positive ion first and the negative ion second. When we are given that name, we must know the appropriate charge on each ion in order to put together a correct formula.<\/p>\n<h3>Binary Ionic Compounds<\/h3>\n<p>A <abbr title=\"A compound made of one metal element and one nonmetal element\">binary ionic<\/abbr> compound contains only <abbr title=\"A single atom that has a charge due to the gain or loss of electrons\">monatomic ions<\/abbr>. Positive monatomic ions have the same name as the element from which they come. Negative monatomic ions are named by modifying the name of the element so that it ends with \u2013ide. Many of the elements in the s block and the p block of the Periodic Table follow the Octet Rule when bonding ionically, so their charges are fairly easy to predict. The table below summarizes these charges:<\/p>\n<table>\n<thead>\n<tr>\n<th>Group<\/th>\n<th>Charge<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Group 1A<\/td>\n<td>1+<\/td>\n<\/tr>\n<tr>\n<td>Group 2A<\/td>\n<td>2+<\/td>\n<\/tr>\n<tr>\n<td>Group 3A<\/td>\n<td>3+<\/td>\n<\/tr>\n<tr>\n<td>Group 4A<\/td>\n<td>no ion formed<\/td>\n<\/tr>\n<tr>\n<td>Group 5A<\/td>\n<td>3-<\/td>\n<\/tr>\n<tr>\n<td>Group 6A<\/td>\n<td>2-<\/td>\n<\/tr>\n<tr>\n<td>Group 7A<\/td>\n<td>1-<\/td>\n<\/tr>\n<tr>\n<td>Group 8A<\/td>\n<td>no ion formed<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Armed with this information, we should be able to write names and formulas for many simple binary ionic compounds. For example, a compound made from calcium and bromine would contain Ca<sup>2+<\/sup> ions and Br<sup>1\u2212 <\/sup>ions. We would name this calcium bromide. To make the net charge of the compound zero, we would need two bromide ions for every calcium ion, so the formula would be CaBr<sub>2<\/sub><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following gives the correct name and formula for a compound made of sodium and phosphorus?<\/p>\n<ol>\n<li>Na<sub>3<\/sub>P; trisodium phosphide<\/li>\n<li>NaP; sodium phosphide<\/li>\n<li>Na<sub>3<\/sub>P; sodium phosphide<\/li>\n<li>NaP; sodium phosphorus<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C. Choices B and D have an incorrect formula. Three Na+ ions are needed for every P<sup>3\u2212 <\/sup>.ion in order for the charges to add up to zero. Choices A and C both have the correct formula, but choice A uses the prefix tri, which is incorrect for an ionic compound.<\/p>\n<\/section>\n<h3>Ions With More Than One Possible Charge<\/h3>\n<p>Many transition metals, as well as several metals near the bottom of the p block (most commonly Sn and Pb), are able to form at least two different ions. Iron, for example, can form Fe<sup>2+<\/sup> and Fe<sup>3+<\/sup>. If we were to name a compound iron sulfide, we would have no idea which ion was in the compound in question. This would then prevent us from being able to write the formula.<\/p>\n<p>We solve this dilemma by using Roman numerals in parentheses as part of the ion\u2019s name. Thus, the name of Fe<sup>2+<\/sup> is not just iron, but iron(II). Fe<sup>3+<\/sup> is named iron(III). Iron(II) sulfide is made from Fe<sup>2+<\/sup> and S<sup>2\u2212<\/sup> ions, so its formula is FeS. Iron(III) sulfide contains Fe<sup>3+<\/sup> and S<sup>2\u2212<\/sup> ions, so its formula is Fe<sub>2<\/sub>S<sub>3<\/sub>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following is a correct pairing of a name and formula?<\/p>\n<ol>\n<li>Tin(IV) chloride; Sn<sub>4<\/sub>Cl<\/li>\n<li>Tin(IV) chloride; SnCl<sub>2<\/sub><\/li>\n<li>Tin(II) chloride; Sn<sub>2<\/sub>Cl<\/li>\n<li>Tin(II) chloride; SnCl<sub>2<\/sub><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Answer The correct answer is D. The tin(II) ion is Sn<sup>2+<\/sup>and the chloride ion is Cl<sup>1\u2212<\/sup>. One Sn<sup>2+<\/sup> is needed for every two Cl<sup>1\u2212<\/sup> ions, so the correct formula is SnCl<sub>2<\/sub>. The correct formula for tin(IV) chloride would be SnCl<sub>4<\/sub>, which does not appear in either choice A or B.<\/p>\n<\/section>\n<h3>Ionic Compounds Containing Polyatomic Ions<\/h3>\n<p>Many ionic compounds contain polyatomic ions. These compounds follow the same rules that we have seen before, with only a few minor differences. Most polyatomic ions are negative (ammonium, NH<sub>4<\/sub><sup>+<\/sup>, is the most common exception), but their names do not usually end with \u2013ide. Most polyatomic ions have names that end with \u2013ite or \u2013ate. Sulfide is simply a sulfur atom with a charge\u2014S<sup>2\u2212<\/sup>. Sulfate, on the other hand, is a sulfur atom and several other atoms together that have a charge\u2014SO<sub>4<\/sub><sup>2\u2212<\/sup>.<\/p>\n<p>The most common polyatomic ions contain an atom along with one or more oxygen atoms. The ion whose name is the element name modified so that it ends in \u2013ate will be considered our \u201cbase\u201d ion. For example, if we modify \u201cnitrogen\u201d so that it ends with \u2013ate, we get \u201cnitrate.\u201d So the nitrate ion, NO<sub>3<\/sub><sup>1\u2212<\/sup>, will be considered our base ion. Unfortunately, there is no hard and fast rule regarding how many oxygen atoms are in the base ion, although it is usually three or four. Once the base ion is known, however, there is a pattern to the names of the other ions made with that element. The following table shows this pattern and uses chlorine (base ion of chlorate, ClO<sub>3<\/sub><sup>1\u2212<\/sup>) as an example.<\/p>\n<table>\n<tbody>\n<tr>\n<td>base ion minus two oxygens<\/td>\n<td><em>hypo<\/em>\u2014root\u2014<em>ite<\/em><\/td>\n<td>hypochlorite, ClO<sup>1\u2212<\/sup><\/td>\n<\/tr>\n<tr>\n<td>base ion minus one oxygen<\/td>\n<td>root\u2014<em>ite<\/em><\/td>\n<td>chlorite, ClO<sub>2<\/sub><sup>1\u2212<\/sup><\/td>\n<\/tr>\n<tr>\n<td>base ion<\/td>\n<td>root\u2014ate<\/td>\n<td>chlorate, ClO<sub>3<\/sub><sup>1\u2212<\/sup><\/td>\n<\/tr>\n<tr>\n<td>base ion plus one oxygen<\/td>\n<td><em>per<\/em>\u2014root\u2014<em>ate<\/em><\/td>\n<td>perchlorate, ClO<sub>4<\/sub><sup>1\u2212<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>By following this pattern, we should be able to see that since NO<sub>3<\/sub><sup>1\u2212<\/sup> is the nitr<u>ate<\/u> ion, then NO<sub>2<\/sub><sup>1\u2212<\/sup> is the nitr<u>ite<\/u> ion.<\/p>\n<p>The only other difference when using polyatomic ions concerns the use of subscripts in the compound\u2019s formula. When a formula requires more than one of a polyatomic ion, we must put the ion formula in parentheses and place the subscript outside of the closing parenthesis. For example, calcium nitrate requires one Ca<sup>2+<\/sup> ion and two NO<sub>3<\/sub><sup>1\u2212<\/sup> ions; therefore, its formula is Ca(NO<sub>3<\/sub>)<sub>2<\/sub>.<\/p>\n<section class=\"question\">\n<h4>Question 1<\/h4>\n<p>Which of the following is the correct formula for copper(I) sulfate?<\/p>\n<ol>\n<li>Cu<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>Cu<sub>2<\/sub>SO<sub>3<\/sub><\/li>\n<li>Cu(SO<sub>4<\/sub>)<sub>2<\/sub><\/li>\n<li>Cu<sub>2<\/sub>S<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A. The correct formula for the sulfate ion is SO<sub>4<\/sub><sup>2\u2013<\/sup>, ruling out choices B (copper(I) sulfite) and D (copper(I) sulfide). Choice C does not have a ratio of ions whose charges add up to zero. Since copper(I) is Cu<sup>+<\/sup>, the correct formula is Cu<sub>2<\/sub>SO<sub>4<\/sub>.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 2<\/h4>\n<p>Which of the following is the correct name for SF<sub>6<\/sub>?<\/p>\n<ol>\n<li>Sulfur(VI) fluoride<\/li>\n<li>Sulfur hexafluoride<\/li>\n<li>Sulfur hexafluorine<\/li>\n<li>Monosulfur hexafluorine<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice B is correct. Sulfur and fluorine are nonmetals that form a molecular compound, so we must use prefixes in the name, eliminating choice A. Choice C is incorrect because fluorine has not been modified to end with \u2013ide. Choice D is incorrect for the same reason and because the prefix mono should be omitted as there is only one sulfur atom in the molecule.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 3<\/h4>\n<p>Which of the following is the correct name for MnO<sub>2<\/sub>?<\/p>\n<ol>\n<li>Manganese oxide<\/li>\n<li>Manganese dioxide<\/li>\n<li>Manganese peroxide<\/li>\n<li>Manganese(IV) oxide<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice D is correct. Like most transition metals, manganese can have several different charges; therefore, the name must use a Roman numeral to tell which ion is present in this compound. Only choice D uses a Roman numeral.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 4<\/h4>\n<p>Which of the following is the correct name for Ca(ClO)<sub>2<\/sub>?<\/p>\n<ol>\n<li>Calcium dichloride<\/li>\n<li>Calcium chloride<\/li>\n<li>Calcium hypochlorite<\/li>\n<li>Calcium chlorate<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice C is correct. Choice A is incorrect because this compound is ionic, so prefixes should not be used. ClO<sup>1\u2013<\/sup> is a polyatomic ion so it does not end with \u2013ide, ruling out choice B. ClO<sup>1\u2013<\/sup> is not the base ion so it cannot end with \u2013ate, ruling out choice D. Because it has two oxygen atoms less than the base ion, it should begin with hypo- and end with \u2013ite, as in choice C.<\/p>\n<\/section>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/comparison-of-properties\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-naming-and-structure\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/acids-and-oxides\">Next Lesson \u27a1<\/a><\/p>\n<\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Chemical Naming and Structure: Writing Names and Formulas Objective In this lesson, we will review how oxidation numbers are assigned. We will also review how to write chemical names and formulas for molecular compounds, ionic compounds, and acids. Previously we covered&#8230; Ionic compounds are held together by strong ionic [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-247","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/247","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=247"}],"version-history":[{"count":19,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/247\/revisions"}],"predecessor-version":[{"id":272,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/247\/revisions\/272"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=247"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}